, 4 min, 778 words
A common stumbling block in an introductory physics class is physics factor problems. They force you to work algebraically, and as a result they can confuse students. I've been helping a cousin with physics recently. He's brilliant at the physics, but can sometimes struggle with figuring out what algebra he needs to do. We talked recently about several "factor problems," and I ended up writing up the procedure I use.
Here's an example:
A coin is on a turntable, at radius $r$ from the center and spinning at constant speed $v$. If the coefficient of friction between the turntable and coin $\mu$ is doubled, how does the maximum rotation speed at which the coin will remain on the turntable change?
Not relevant here are free body diagrams or the actual "just before slipping" condition. All I want to talk about is how to navigate a problem that gives you one factor and asks you to find another one.
This steps requires you to do the actual physics of the problem. Here I would draw a free body diagram (hint: gravity, normal force, friction). Then I'd think about circular motion and centripetal acceleration. Here friction is the centripetal force. So after a bit of physics, the relevant equation turns out to be:
$\frac{mv^2}{r} = \mu m g$
Now it's time to use the information the problem gave us. We're going to create two new variables.
The first variable is the one that the problem tells us to change: $\mu_2=2\mu_1$
The second is what we're solving for: $v_2=(??)v_1$
Our goal now is to find the factor between the old and new variables of the "solving for" variable. In other words, solve for the question marks.
Here, that means switching $\mu$ for $\mu_1$ and $v$ for $v_2$. We get:
$$\frac{m{v_1}^2}{r}=\mu_1 m g$$
Pretty straightforward, right? We just added a couple subscripts. Just wait 'til the next step.
$$\frac{m{v_2}^2}{r}=\mu_2 m g$$
Wow, that sure was exciting...
In step (2), we defined $\mu_2=2\mu_1$. Let's plug that in up above.
$$\frac{m{v_2}^2}{r}=(2 \mu_1) m g$$
This is often the hardest step. In that equation in step (5), we see something that looks a lot like step (3), right? In this case, the similar expression is $\mu_1 m g$ and $(2 \mu_1) m g$. We can factor this:
$$\frac{m{v_2}^2}{r}=2 (\mu_1 m g)$$
That may not seem useful yet, but it will be!
From up in (3), we know $\mu_1 m g = \frac{m{v_1}^2}{r}$. So let's plug that in!
$$\frac{m{v_2}^2}{r}=2\left(\frac{m{v_1}^2}{r}\right)$$
I know, I know. This is just looking scarier and scarier. But it's about to lose some of that terror!
Here we can divide both sides by $m$ and multiply by $r$. That gives us:
$${v_2}^2=2{v_1}^2$$
And finally...(drumroll please)
Just take the square root of both sides here, and you get:
$$v_2 = \sqrt{2}v_1$$
In other words, if we double the coefficient of friction, the max speed of the coin where it won't slip increases by a factor of root two.
Phew!
In general, once you have practice you won't need to write out every step of the procedure for physics factor problems. Here's what it would look like if I were writing it up in detail:
As discussed above, the relevant equation relating velocity and the coefficient of friction is
\[\frac{mv^2}{r}=\mu m g.\]
Now if we double the coefficient of friction ($\mu_2=2\mu$), then we'll get a new velocity $v_2$ given by
\[\frac{m{v_2}^2}{r}=(2\mu) m g = 2(\mu m g).\]
Plugging in the equation above, we see
\[\frac{m{v_2}^2}{r}=2(\mu m g) = 2\left(\frac{mv^2}{r}\right),\]
which we can solve for $v_2$:
\[{v_2}^2=2v^2,\]
and thus $v_2=\sqrt{2}v$. In other words, doubling the coefficient of friction will increase the max speed at which the coin can revolve without slipping by a factor of $\sqrt2$.